In a diagram of electric field lines, q1 has 9 lines going into it and q2 has 27 lines going out of it. If one of
the charges is -40 mC, which of the following could be the other one?
You offer no choices. The total E-field flux leaving a charge is proportional to the charge. (Gauss' law). The charge with field lines going in is negative. The other charge (with field lines going out) must be positive, and three times the magnitude, or 120 mC
Well, it seems like q1 is a bit shy, with all those electric field lines going into it. And q2, on the other hand, seems to be quite popular with all those lines going out of it. But hey, no judgment here!
Now, let's see if we can find a match for q1. Since it has 9 lines going into it, it must have a weaker charge compared to q2. If we assume q2 is positive, then q1 should be negative to attract those electric field lines.
So, if q1 is -40 mC, well, that's quite a negative charge! And as for q2, it must be positive and have a stronger charge than q1 to send out those 27 lines.
So, the other charge could be a positive charge, but we don't know its exact value. It just has to be stronger than -40 mC so that it can send all those 27 electric field lines out towards q1.
Hope that enlightens you a bit, or should I say, electrifies your understanding!
To determine the charge of the other object, we need to understand the relationship between the number of electric field lines and the magnitude of the charge.
According to Gauss's Law, the number of electric field lines that pass through a closed surface is directly proportional to the charge enclosed by that surface.
In this case, we have q1 with 9 lines going into it and q2 with 27 lines going out of it. This indicates that the magnitude of charge enclosed by q1 is three times that of q2.
Let's denote the magnitude of charge of q2 as Q2. Therefore, the magnitude of charge of q1 is 9 * Q2.
Now, we can consider the given charge of -40 mC. Since charges can be positive or negative, we can have two possible combinations:
1. q1 = -40 mC and q2 = -13.33 mC
In this case, the magnitude of charge of q2 would be -40 mC / 9 = -4.44 mC, rounded to two decimal places.
2. q1 = 40 mC and q2 = 13.33 mC
In this case, the magnitude of charge of q2 would be 40 mC / 9 = 4.44 mC, rounded to two decimal places.
So, the charge of the other object (q2) could be either -13.33 mC or 13.33 mC.
To determine the possible value of the other charge, let's first understand the concept of electric field lines.
Electric field lines represent the direction and relative strength of the electric field around a charged object. The number of electric field lines is proportional to the magnitude of the charge. Electromagnetic theory states that electric field lines start from positive charges and end on negative charges.
Here, we are given that charge q1 has 9 lines going into it, and charge q2 has 27 lines going out of it.
Since electric field lines start from positive charges, we can conclude that q1 is positive. Similarly, since electric field lines end on negative charges, we can conclude that q2 is negative.
Now, we are given that one of the charges is -40 mC (negative charge). Let's determine the magnitude of the other charge.
Since q1 has 9 lines going into it, it represents a smaller magnitude of charge compared to q2, which has 27 lines going out of it. The ratio of electric field lines can be expressed as:
q1 / q2 = 9 / 27
Simplifying the ratio:
q1 / q2 = 1 / 3
To find the possible value for the other charge, we can set up an equation using the known quantity:
-40 mC / q2 = 1 / 3
Simplifying the equation:
-120 mC = q2
Therefore, the other charge q2 could be -120 mC.